Answer:
SI unit of k (spring constant) = N/m
Explanation:
We have expression for force in a spring extended by x m given by
F = kx
Where k is the spring constant value.
Taking units on both sides
Unit of F = Unit of k x Unit of x
N = Unit of k x m
Unit of k = N/m
SI unit of k (spring constant) = N/m
Note: The complete question is attached as a file to this solution. The parallel plate mentioned can be seen in this picture attached.
Answer:
E = 225 N/C
Explanation:
Note: At any point on the parallel plates of a capacitor, the electric field is uniform and equal.
Therefore, Electric field at x = 14 cm equals the electric field at x = 7 cm
V(x) = 31.5 Volts
x = 14 cm = 0.14 m
The magnitude of the electric field at any point between the parallel plate of the capacitor is given by the equation:
E = V(x)/d
E(x = 0.14) = 31.5/0.14
E(x=0.14) = 225 N/C
E(x=0.14) = E(x=0.07) = 225 N/C
D.<span>moving at a constant speed of 150 km/h
Acceleration is the change in velocity.
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Answer:
Biological system is one of the major causes of oscillation due to sensitive negative feedback loops. For instance, imagine a father teaching his son how to drive, the teen is trying to keep the car in the centre lane and his father tell him to go right or go left as the case may be. This is a example of a negative feedback loop of a biological system. If the father's sensitivity to the car's position on the road is reasonable, the car will travel in a fairly straight line down the centre of the road. On the other hand, what happens if the father raise his voice at the son "go right" or when the car drifts a bit to the left? The startled the son will over correct, taking the car too far to the right. The father will then starts yelling "go left" then the boy will over correct again and the car will definitely oscillate back and forth. A scenario that indicates the behavior of a car driver under a very steep feedback control mechanism. Since the driver over corrects in each direction. Therefore causes oscillations.
Explanation:
Answer:
Explanation:
The blocks are moving in the x direction
Given that,
m1=3.03kg
m2=6.40kg
After a force F is applied to the two masses they move with an acceleration of 2.7m/s²
Newton's Second Law:
ΣF = Σ(ma) = m1a1 + m2a2
Therefore
But in this case a1 =a2=2.7m/s², because they are connected by a string and they are accelerating at the same speed.
F=(m1+m2)a
F=(3.03+6.40)2.7
F=9.43 ×2.7
F=25.46N.
b. The contact force be blocks
F2= m2a
F2 is the force between the blocks
F2=6.4×2.7
F2= 17.28N
c. Fnet on block 1
It is the net force pulling the two blocks forward minus the force between the block 1 and 2
F1=F-F2
F=25.46-17.28
F=8.18N
The net force on block 1 is 8.18N